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## Algebra 1

### Course: Algebra 1 > Unit 13

Lesson 5: Factoring quadratics intro- Factoring quadratics as (x+a)(x+b)
- Factoring quadratics: leading coefficient = 1
- Factoring quadratics as (x+a)(x+b) (example 2)
- More examples of factoring quadratics as (x+a)(x+b)
- Factoring quadratics intro
- Factoring quadratics with a common factor
- Factoring completely with a common factor
- Factoring quadratics with a common factor
- Factoring simple quadratics review

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# Factoring quadratics as (x+a)(x+b) (example 2)

Sal factors x^2-14x+40 as (x-4)(x-10) and x^2-x-12 as (x+3)(x-4). Created by Sal Khan.

## Want to join the conversation?

- I didn't get0:58(15 votes)
- The product of two numbers just means what you get when the two numbers are multiplied together. He's looking for two numbers a and b such that a + b (their sum) is -14 and ab (their product) is 40.(27 votes)

- Sal says that this is guess work, but:

a+b = -14 and ab = 40

looks like a very obvious 'systems of equations' problem. When I tried it though things got messy very quickly. Neither substitution nor elimination worked. Can anyone explain why?(10 votes)- This arises from a quadratic, and if you try to solve for either a or b, you will get a quadratic again. The best procedure is just to test factors of 40 until you realize that -4 and -10 are legitimate solutions to this system.(6 votes)

- why do they call it a quadratic?(5 votes)
- Since the highest exponent is a 2.
`x^2 + 2x - 3`

.(11 votes)

- how can I deeply learn this some more. also where in my life would I have to use this besides math class lol!! some examples might help(8 votes)
- yo tobias i ask the same question everyday, yet everyday i tell myself i need this credit!(3 votes)

- What is the purpose in Factoring?(2 votes)
- It can help us find roots of polynomials, and it can help us simplify rational expressions.(14 votes)

- I apologize for not being very specific, but does anyone know a faster way to factor? This is the same way we learned it in school and I would like to know some formulas, etc., to factor polynomials like these. Thanks!!(2 votes)
- I don't think there are any formulas to factor polynomials, but there always is the quadratic formula. This is any easy way of finding roots (x-intercepts) of a quadratic equation by just plugging values into an equation.(7 votes)

- does the order of the (x )(x ) matter if they are the same coefficients with the correct sign? say the solution is (x-2)(x+3). would it matter if it was (x+3)(x-2)?(4 votes)
- The order does not matter because the commutative property of multiplication applies. This is just like 2(3) is the same as 3(2).

Hope this helps.(3 votes)

- What if you have something that can't be factored?(2 votes)
- Then you leave it as it is. For exmple
`x^2 + 6x + 23`

can not be factored because no two numbers multiplied together will equal 23.(6 votes)

- how would you factor a value for x^2 where the coefficient is greater than 1?(2 votes)
- Simply put, we would just change the value that we need a and b to multiply to. Currently, we always try to get a and b to multiply to whatever the last term in our quadratic is. But really what we're looking for is for our a and b to multiply to the coefficient of the first term and the value of the last term. Here's a better example.

All quadratics are written in the form:

ax^2 + bx + c

(In this case, a and b have**no**relation to the a and b that Sal is talking about for factoring.)

Based on this equation, we want our two factors to multiply to a*c.

For example, if we have the equation:

4x^2 + 9x + 10

Then we want our a and b to multiply to 4*10.

Hope that helps a little!(5 votes)

- (3x + 4y)^2 + 2(3x +4y)(3x-4y) + (3x-4y)^2

Can this be factorised?

How?(2 votes)- There are two ways to handle this. You can substitute
`v = 3x+4y`

and`w = 3x - 4y`

, then you have`v² + 2 v w + w² = (v + w)(v + w) = (v + w)²`

Now you can resubstitute your v and w`(v + w)² = (3x + 4y + 3x - 4y)² = (6x)² = 36x²`

You could also get there by expanding your terms first`(3x + 4y)² + 2(3x + 4y)(3x - 4y) + (3x - 4y)²`

= (9x² + 24xy + 16y²) + (18x² - 32y²) + (9x² - 24xy + 16y²)

= 9x² + 18x² + 9x² + 24xy - 24xy + 16y² -32y² + 16 y²

= 36x²(5 votes)

## Video transcript

To better understand how
we can factor second degree expressions like this, I'm going
to go through some examples. We'll factor this expression and
we'll factor this expression. And hopefully it'll
give you a background on how you could generally
factor expressions like this. And to think about
it, let's think about what happens if I were
to multiply x plus something times x plus something else. Well, if I were to multiply
this out, what do I get? Well, you're going to get
x squared plus ax plus bx, which is the same thing as
a plus bx plus a times b. So if you wanted to go
from this form, which is what we have in these
two examples, back to this, you really just have
to think about well, what's our coefficient
on our x term, and can I figure out two numbers
that when I take their sum, are equal to that coefficient,
and what's my constant term, and can I think of two numbers,
those same two numbers, that when I take the product
equal that constant term? So let's do that over here. If we look at our
coefficient on x, can we think of
an a plus ab that is equal to that
number negative 14? And can we think
of the same a and b that if we were to
take its product, it would be equal to 40? So what's an a and a b
that would work over here? Well, let's think about
this a little bit. If I have 4 times 10
is 40, but 4 plus 10 is equal to positive 14. So that wouldn't quite work. What happens if we make
them both negative? If we have negative
4 plus negative 10, well that's going to be
equal to negative 14. And negative 4 times
negative 10 is equal to 40. The fact that this number
right over here is positive, this number right
over here is positive, tells you that these are
going to be the same sign. If this number right
over here was negative, then we would have
different signs. And so if you have
2 numbers that are going to be the same
sign and they add up to a negative number,
then that tells you that they're both
going to be negative. So just going back
to this, we know that a is going to be negative
4, b is equal to negative 10, and we are done factoring it. We can factor this expression
as x plus negative 4 times x plus negative 10. Or another way to
write that, that's x minus 4 times x minus 10. Now let's do the
same thing over here. Can we think of
an a plus b that's equal to the coefficient
on the x term? Well, the coefficient
on the x term here is essentially
negative 1 times x. So we could say the
coefficient is negative 1. And can we think of
an a times b where it's going to be
equal to negative 12? Well, let's think about
this a little bit. The product of the 2
numbers is negative, so that means that they
have different signs. So one will be positive
and one will be negative. And so when I add the two
together, I get to negative 1. Well, just think about the
factors of negative 12. Well, what about if one is 3
and maybe one is negative 4. Well, that seems to work. And you really just have
to try these numbers out. If a is 3 plus
negative 4, that indeed turns out to be negative 1. And if we have 3
times negative 4, that indeed is equal
to negative 12. So that seems to work out. And it's really a matter
of trial and error. You could try negative
3 plus 4, but then that wouldn't have worked
out over here. You could have
tried two and six, but that wouldn't have
worked out on this number. Or 2 and negative
6, you wouldn't have gotten the sum to
be equal to negative 1. But now that we've figured
out what the a and b are, what is this
expression factored? Well, it's going to be x plus
3 times x plus negative 4, or we could say x minus 4.